Question

In a hot water heating system there is a cylindrical pipe of length 28m and diameter 5 cm. Find the total radiating surface in the system system.​

Answer 1

Diameter = 5 cm
Radius = Diameter ÷2=5cm÷2=2.5cm
length of pipe =28m=2800cm
Total Radiating Surface of pipe = Curved surface area + 2 × Area of circular crossection
=2πrh+2πr
2
=2πr(h+r)=2×
7
22
.
.
.
plz mark me brainliest

​
×2.5(2800+2.5)=44039.28cm
2

Answer 2

$\huge\mathbb\pink{Hey \: Mate!!}$

${\huge{\red{\underline{\overline{\mathit{\red{ANSWER↓}}}}}}}$

h = 28m

2r=5cm

$\therefore$ r = $\large\dfrac{5}{2}$ cm = $\large\dfrac{5}{2×100}$

= $\large\dfrac{5}{200}$ m = $\large\dfrac{1}{40}$ m

$\therefore$ Total radiating surface in the system

= ${\small{\boxed{\sf{\pink{2πrh}}}}}$

= 2 × $\large\dfrac{27}{7}$ × $\large\dfrac{1}{40}$ × 28 = ${\underline{\underline{4.4m²}}}$

The area of the radiating surface of system is 4.4 m²

$\large\underline\mathbb\red{HOPE \: THIS \: HELPS \: U}$

$\large\fbox\colorbox{pink}{Mark \: as Brainliest}$